As for the second example, suppose we have the three digit number ABC with smaller reverse CBA. Since this is smaller, it must be that C < A. Subtracting the smaller from the larger gives us (A - C)0(C - A). Except C is less than A, so there must be a borrow for the last digit's subtraction: we actually get (A - C - 1)9(10 + C - A). Finally, adding this to its reverse gives 9(9 + 9)9 = 1089.

Here's an easier version of the trick that's too obvious and lots of people would see right through it:

Take the first digit of your age, multiply it by ten, then add one, then add the second digit of your age and tell me the result. The result is 25? Aha! I know that your age is really 24.

To make it a little more sneaky, they added six instead of one, and they broke up the multiplying by ten into two steps, first asking you to multiply by five and then later multiply by two. And the icing on the cake is that instead of asking you to add six at the end, they had you add three just before you multiply by two, which has the same effect.

If AB is your age, i.e. A is the first digit of your age and B is the second digit...

"Multiply A by ten then add B" gives you the exact same answer as
"Multiply A by ten then add B then add six then subtract six" which is the same as
"Multiply A by ten then add six then add B then subtract six" which is the same as
"Multiply A by five then multiply again by two then add six then add B then subtract six" which is the same as
"Multiply A by five then add three then multiply by two then add B then subtract six".

I was showing my steps in the most compact way possible. To avoid a tedious and long post.

Understood; and in this context, I'm inclined to forgive you. In an educational context, however, you should bend over backward to avoid the "running equals sign," since thinking of = as "gives" or "execute" can be a serious stumbling block to kids' understanding of math. See "Studentsí Understanding Of The Equal Sign Not Equal" (or more discussion here, here, or here.

What do you use instead of equals sign for "execute" in such a simple case as above? (Although I would have said "gives," were I orally stepping through the symbols).

I say "simple case" because no doubt simple opertions/demonstrations in other formalist endeavors are governed by unique symbologies.

One symbol commonly used is "=>" which is pronounced "yields" or "gives."

Quote:

Originally Posted by aceplace57

Thank you for explaining the misdirection sbunny8.

I see now how clever they were. Adding steps to hide what they were really doing.

Understanding how it's designed makes me appreciate the trick much more.

Actually it proves how dirt-simple these puzzles are. Now that you know the tricks you can make up your own private version and baffle your friends.

As has been said, it all starts from the fact that all 2 digit numbers are actually in the form of 10a+b. So as long as you hide a multiply by 10 and a single addition in amongst a bunch of do-nothing steps you've got a working puzzle. e.g. take a digit, add 3 then double the result, then subtract 6, then divide by 2. Magically the same digit pops out. etc.

I see now how clever they were. Adding steps to hide what they were really doing.

Understanding how it's designed makes me appreciate the trick much more.

To make the trick less obvious, they should reverse the digits.

Pick a number between 11 and 99.
Take the second digit and multiply it by five.
Add three.
Multiply it by two.
Add the first digit from your original number.
What number did you come up with?
Your original number is...

All they do is take the number you gave them, subtract six, and reverse the order of the digits. But because of that reversal the trick is less obvious.

With the original trick

17 => 11
30 => 24
44 => 38
57 => 51
66 => 60
82 => 76
105 => 99

Even a person with limited arithmetic skills might spot such a simple pattern.

I can see this trick going horribly wrong. You're 39! No I'm not! Then it becomes obvious the mental arithmetic slipped a gear.

Yeah. I remember when I was in 5th grade, trying a similar trick on my cousin, who is two years older:

"Think of a number. Double it. Add 10. Divide by 2. Subtract the number you first thought of. Your answer is 5."

He laughed hysterically, making fun of me because the trick didn't work. He had just learned about negative numbers in school, so he started with -1. He said that I wasn't so smart, because my trick doesn't work with negatives.

Two years later, I learned about negative numbers myself, and discovered that the trick works fine with them. I guess he didn't understand negatives as well as he thought. (I never did bring it up again, though.)

As has been said, it all starts from the fact that all 2 digit numbers are actually in the form of 10a+b. So as long as you hide a multiply by 10 and a single addition in amongst a bunch of do-nothing steps you've got a working puzzle. e.g. take a digit, add 3 then double the result, then subtract 6, then divide by 2. Magically the same digit pops out. etc.

Well, these are tricks to impress children, primarily, right? Illustrations of algebra to entice those who haven't yet thought about it.

Incidentally, even if it wasn't so smooth, you would imagine any series of steps of this sort could be undone easily enough: take your age, add 2, multiply it by 8, subtract 3, square it, reverse the digits, and tell me what you got. Ok, now I'll deduce your age: I'll reverse the digits, take the square root, add 3, divide by 8, and subtract 2. That sort of thing.

(Hell, since there's only so many ages someone (and particularly a child) might reasonably have, I may as well just invent any old random complicated but injective function from input ages to output whatevers and just memorize the mapping back from outputs to inputs.)

Nines lead to some real fun tricks. One I show my students goes like this:

Think of a four-digit number. Don't make it something easy like 5555; make it hard.
Now scramble the digits to make a new number.
Subtract the smaller number from the larger number. You'll probably end up with a four-digit number; if not, tell me the number of digits so I can direct the rest of the trick. (If the person's a prick and gets a one-digit number, you gotta tell them they're a prick and to rescramble; otherwise the adjustments should be obvious).
Take that new four-digit number and scramble it to get a new number.
Choose one of the digits in that new number and circle it. Don't circle a zero, that's already a circle. (If that line doesn't set off alarm bells in your head, you don't pay much attention to magic tricks). (Don't say that last part).
Take the three uncircled digits, and scramble them to make a new number.
Tell me what that new number is, and I'll tell you what number you circled.

***

As for the 1089 trick, I do it slightly differently: I show kids my new speedreading technique (riffling quickly through the pages of a book) and tell them that I've memorized the book completely. Then we do the addition/subtraction shenanigans--if they got a two-digit number after subtraction, it's 99, so I tell them to reverse and add twice instead of once. I tell them to look at the first two digits of their number to choose a page in the book, the third digit to choose the line number on that page, and the final digit to choose the word on that line. They tell me they're on the tenth page, line eight, word nine. I've memorized that sentence, so I make a big show of muttering it to my breath and counting words: "Hmm, Frodo's at the party, he's talking with Samwise, let's see, 'after, the, party, is, finished, would, you, be, so'--the word is 'so,' right?"

It's worth noting that there's two basic modes for these tricks: creating an injective [and indeed easily invertible] function, as in trick 1 in this thread (where the trickster can deduce the input from the trickee telling them the output), and creating a constant function, as in trick 2 in this thread (where the trickster can deduce the output even without knowing the trickee's input).

The latter form, though it sounds so unimpressive written out as just "this function turns out to be constant", connects to fond memories for me, as it was what they used to use at the end of some episodes of Square One TV, my first exposure to these tricks as a child (obviously, on TV, the interactivity required for the former couldn't be employed).

Speaking of four-digit numbers and constant functions, here's a trick:

Take any four digit number whose digits aren't all the same (e.g., 7274), rearrange its four digits into order (e.g., 2477), and subtract this from its reverse (e.g., 7742 - 2477 = 5265).

Keep repeating this process till you get the same value twice in a row (so, in this example, 5265 then becomes 6552 - 2556 = 3996; this in turn becomes 9963 - 3699 = 6264; and one can continue in this fashion a couple more times before getting a repeat).

The person says 38 and you tell them they are 32. By subtracting 6. ***applause***

First digit times five... well in the two digit number, thats the number of tens, and the number of ones is being added on again later, so "times five" is really "divide by two".. 3 times five is 30 divide two... see ?

So then add 3, and then double . You can move the 3 to be after the double.. well the "add three" was doubled , so thats really add six.. which is then cancelled by subtract 6.

Then add in the ones column, and subtract six ...

basically the add 6 is disguised as "halve, add 3 and double", but the halve is disguised as tens column times 5...

Here's a trick I like that looks somewhat similar to the previous examples, yet whose explanation may elude some, and which at any rate connects to interesting deeper math. Alas, its one fault is that most people aren't familiar with how to carry out the required large exponentiations. But, here, I'll link you to a calculator you can use for this purpose.

Pick any three-digit even number you like. Write a 1 after it, and then write in an exponent of 1 followed by three 0s. (For example, if your original number was 712, it would become 71211000)

Carry out this exponentiation. The result will have lots of digits; in particular, it will end in three 0s followed by a 1. Get rid of those. The result will still be huge; cut it down to size by keeping only the last three digits at this point. (For example, 71211000 is very large, and its last digits are "...2204035920001". Getting rid of the 0001 chunk and then keeping only the last three digits, we end up with 592). This value will be even, so halve it, and then tell me what you get. (In this case, that would be 592/2 = 296).

At this point, armed with nothing else, I will readily recover your original number. In fact, the most surprising thing is the way in which I will recover your original number: I raise the "magic number" 8221 to what you gave me, chop off the last digit, and then keep only the last three digits. [E.g., 8221296 has last digits "...733817121". Chopping off the last 1 and then keeping only the last three three digits, I get "712" back, just as you started with].

The challenge to you (after you are able to actually try out enough examples to assure yourself this does indeed work) is to explain: why does this work?

(In particular, you should know I can do this trick for any number of digits, in any base, just changing the "magic number" accordingly.)

In RSA-style encryption, you first raise a secret to one power, then undo by raising to another power. But, here, we first raise the secret to some power, then undo by raising another base to that. So not quite the same.

Pick any three digit number you like. Write a 1 after it and then raise it to the 63rd power. Chop off the last digit from the result, and then tell me the last three digits. (For example, if your original number was 712, it would become 712163, whose last digits are ...42419761, and you would tell me 976)

I can now recover your original number by adding back in the final 1 you chopped off, raising it to the 127th power, chopping off the last 1 again, and reporting the last three digits. (For example, 976 becomes 9761127, whose last digits are ...6061297121, and I would report to you 712)

And there are variants of this for any number of digits, any base, and also allowing different pair of exponents, and even allowing different digits to add and chop off, the latter subject to certain coherence conditions. That's an RSA-style trick.

Binomial expansions are certainly good things to look at. I'd advise (1 + 10A)^10. Indeed, I'd advise just playing around and seeing what happens to numbers ending in 1 as you repeatedly raise them to 10th powers. That should make at least some relevant patterns clear, which may spur further thoughts. I'll provide a fuller explanation in a few days if there's still interest.

Last edited by Indistinguishable; 08-30-2016 at 12:11 AM.
Reason: Trying to limit the amount I procrastinate work I should be doing via wasting time on the Internet

I'll also note that the significance of the magic number 8221 is that there is a sense in which e20 = ...8221 (in base ten), and a sense in which the procedure outlined at the beginning of post #37 carries out the operation ln(1 + 10x)/20 in mod 1000 land.

While out process serving I get to guess ages all day long, its fun. Sometime dead on but always within 5 years and usually with 3.

Its hard sometimes because you dont want to insult the women

The weight too...pretty good at that one too! On women I always guess less than what they look like.....if I feel good that day I will do an obviously 30+ woman and guess 25...that really thrills them! Course I have to do anything to get them back in a good mood or in a good mood after I serve them!

Most dont like getting served. Not too many at all slam the door on me!!

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